JEE Advanced 2018 — Paper 1 — Question 32

Given $f(x)=1-2x+e^{x}\int_{0}^{x}e^{-t}f(t)\,dt$. Differentiate using Leibniz rule: $f'(x)=-2+e^{x}\int_{0}^{x}e^{-t}f(t)\,dt + f(x)$. Substitute the original equation: $e^{x}\int_{0}^{x}e^{-t}f(t)\,dt = f(x)-1+2x$. Thus $f'(x) = -2 + (f(x)-1+2x) + f(x) = 2f(x)+2x-3$. So $f'(x)-2f(x)=2x-3$.

This is a linear ODE with integrating factor $e^{-2x}$. Then $\frac{d}{dx}[f(x)e^{-2x}] = (2x-3)e^{-2x}$. Integrate: $f(x)e^{-2x} = \int (2x-3)e^{-2x}\,dx = (1-x)e^{-2x}+C$. Using $f(0)=1$ gives $C=0$, so $f(x)=1-x$. Check options: $f(2)=-1$ so (2,-1) lies on the curve (option B true). Area between $y=1-x$ and $y=\sqrt{1-x^{2}}$ from $x=0$ to $1$ is $\int_{0}^{1}[\sqrt{1-x^{2}}-(1-x)]\,dx = \frac{\pi}{4}-\frac{1}{2} = \frac{\pi-2}{4}$ (option C true).